= Solution
For a steady thin-wire <electric current>, the <Biot-Savart law> is
$$
d\mathbf B(\mathbf r)=\frac{\mu_0I}{4\pi}
\frac{d\boldsymbol\ell'\times(\mathbf r-\mathbf r')}{|\mathbf r-\mathbf r'|^3},
$$
where the source element follows positive current. For a circular loop, source points are $(R\cos\phi,R\sin\phi,0)$ and the observer is $(0,0,z)$. Their <cross product> with $d\boldsymbol\ell'$ has axial component $R^2d\phi$; transverse components integrate to zero. Thus the <on-axis magnetic field of a circular current loop>, multiplied by the number of coincident turns, is
$$
\boxed{B_0(z)=B_z(0,z)=\frac{\mu_0N_1I_1R^2}{2(R^2+z^2)^{3/2}}.}
$$
Take positive current counterclockwise as seen from $+z$. <magnetic field lines> pass upward through the loop and return around its exterior, closing rather than terminating. They fan outward above its centre and converge below it. The original numerical <Biot-Savart law> integration below illustrates these directions; the wire singularities are masked.
\Image[/past-exam-of-the-physics-b-course-of-the-university-of-cambridge/2012/paper-1-coil-field.png]
{title=Original circular-loop <magnetic field lines> in a meridional cross-section}
The <axisymmetric magnetic field near a regular axis> follows from <Gauss's law for magnetism>. In <cylindrical coordinates>, regularity gives $B_\rho(0,z)=0$, while <symmetry> gives $B_z(\rho,z)=B_0(z)+O(\rho^2)$. Integrate $\partial_\rho(\rho B_\rho)=-\rho B_0'(z)$ from the axis:
$$
B_\rho(\rho,z)=-\frac\rho2 B_0'(z)+O(\rho^3),
\qquad
\boxed{B_\rho(\rho,z)\simeq
\frac{3\mu_0N_1I_1R^2\rho z}{4(R^2+z^2)^{5/2}}.}
$$
A formal $C(z)/\rho$ term is excluded by regularity. For $z>0$ the axial field decreases upward, so the radial component is outward, consistent with the sketch.
On the small receiving coil, $d\boldsymbol\ell=a\,d\phi\,\widehat{\boldsymbol\phi}$. The <Lorentz force> $I_2d\boldsymbol\ell\times\mathbf B$ has axial component $-I_2a B_\rho(a,d)d\phi$. Contributions from the axial field cancel radially around the loop. Therefore
$$
\boxed{F_z\simeq-2\pi N_2I_2aB_\rho(a,d)
=-\frac{3\pi\mu_0N_1N_2I_1I_2R^2a^2d}{2(R^2+d^2)^{5/2}}.}
$$
With both currents giving <magnetic dipole moments> along $+z$ and $d>0$, this is attraction. Reversing one current reverses the <force>. The small coil has $m_2=N_2I_2\pi a^2$, so the result is exactly the leading <magnetic dipole force> $F_z=m_2B_0'(d)$. Equivalently $U=-\mathbf m_2\cdot\mathbf B_0$. The approximation requires $a\ll R$ and negligible variation over the receiving coil; corrections begin at relative order $a^2/(R^2+d^2)$.
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